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80 Organic Chemistry Solutions Manual Suggested solution Compound A contains the two reagents combined with the loss of HBr, and the four Hs at 6.74 and 7.18 p.p.m. suggest that the other reagent is attached to the benzene ring. The OMe group is still there (3H singlet at 3.72 p.p.m.) and the new signals are a coupled 6H doublet and 1H septuplet - an isopropyl group. The compound is one of these three isomers. Br or AICI₃ MeO MeO ortho para meta The correct isomer can be determined by the detailed NMR of the aromatic protons. They fall into two groups of 2H each and each group is coupled with the other with a 9 Hz coupling. This fits only with the para isomer. Both ortho and meta isomers have four different aromatic protons. Notice that we did this using the proton NMR only, but we might as well assign the other data. There is no IR as there are no functional groups except an ether. The four signals in the aromatic region of the carbon NMR also reflect the symmetry of the molecule. 7.18 24 H 2H, 1.21 Br 127 or 115 33 6.74 6H, 2H, H 141 24 H MeO AICI₃ 2.83, 1H MeO 153 MeO septuplet 59 para 3.72 para Compound B combines the two reagents with the loss of Me₃Si and the gain of H. Both the infrared and the 13 C NMR show the appearance of a second carbonyl group - the ester (1745 cm⁻¹. 176 p.p.m.) has been joined by an aldehyde or ketone (1730 cm⁻¹, 202 p.p.m.). The proton NMR shows that it is an aldehyde (10.01, 1H, s). There is also a CMe₂ group, but no longer part of an alkene (proton and carbon NMR also show that the alkene has gone) and the OMe group of the ester (4.3, 3H, s). Finally, and very helpfully, there are two open-chain CH₂s linked together (the two triplets with J7 Hz). One of them (2.24 p.p.m.) is next to something and this has to be one of the carbonyl groups as there isn't anything else! So we have: H H 0 Me Me CO₂Me OSIMe₃ CO₂Me CHO 7 ? ? H H The carbonyl group on the chain is in brackets because we already have it as the or the ester - we just don't know which. There are only two ways to join these fragments up. Me Me Me Me or OSiMe₃ CHO CO₂Me Though we mustn't say what 'ought to happen' we can still see that it makes more sense for the group to stay where it is, on a chain of two carbon atoms, than mysteriously move to the other end the molecule next to the two methyl groups. This is not evidence, but we prefer the second structure Real evidence comes from the absence of coupling between the aldehyde proton and anything. If first structure were correct, the aldehyde proton would be a triplet. The second structure is right. 15 1.21 2.24, 2H, t Me Me 6H, S Me Me H H CO₂Me 62 OSiMe₃ 48 CO₂Me CO₂Me 22 202 176 H H 1.8, 2H,